Given the vertices A(-3,-2,0) , B(3,-3,1) , C(5,0,2)…

Given the vertices $A(-3,-2,0)$, $B(3,-3,1)$, $C(5,0,2)$, $D(-1,1,1)$ of the parallelogram $ABCD$, find the angle between its diagonals.

Explanation of Method

The angle between the diagonals of the parallelogram ABCD.

Step 1: It is a parallelogram

\[ \overrightarrow{AB}=(6,-1,1)=\overrightarrow{DC}, \] so $ABCD$ is indeed a parallelogram.

Step 2: The diagonal vectors

\[ \overrightarrow{AC}=(8,2,2),\qquad \overrightarrow{BD}=(-4,4,0), \]

\[ |\overrightarrow{AC}|=\sqrt{72}=6\sqrt2,\qquad |\overrightarrow{BD}|=\sqrt{32}=4\sqrt2. \]

Step 3: The angle

\[ \overrightarrow{AC}\cdot\overrightarrow{BD}=-32+8+0=-24, \]

\[ \cos\angle(\overrightarrow{AC},\overrightarrow{BD})=\frac{-24}{6\sqrt2\cdot4\sqrt2}=\frac{-24}{48}=-\frac12. \]

The vectors make $120^\circ$. Two lines make the acute angle $180^\circ-120^\circ$: \[\varphi=60^\circ\quad(\text{the vectors }\overrightarrow{AC},\overrightarrow{BD}\text{ make }120^\circ)\]

$M$ $A(-3,-2,0)$ $B(3,-3,1)$ $C(5,0,2)$ $D(-1,1,1)$ $O$

$\overrightarrow{AC},\ |AC|=6\sqrt2$
$\overrightarrow{BD},\ |BD|=4\sqrt2$

The diagonals bisect each other at $M(1,-1,1)$; the vectors make $120^\circ$, so the (acute) angle between the diagonals is $60^\circ$. View from above, so the parallelogram is not foreshortened.

Result

\[\boxed{\,\varphi=60^\circ\quad(\text{the vectors }\overrightarrow{AC},\overrightarrow{BD}\text{ make }120^\circ)\,}\]